chemistry i honors
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Chemistry I Honors. Unit 9 : Gases. Objectives #1-4: Introduction to the Kinetic Theory of Gases. The Kinetic Theory Assumptions of the Kinetic Theory Gases are composed of tiny particles that are arranged far apart from each other - PowerPoint PPT PresentationTRANSCRIPT
Chemistry I Honors
Unit 9: Gases
Objectives #1-4: Introduction to the Kinetic Theory of Gases
I. The Kinetic TheoryA. Assumptions of the Kinetic Theory
Gases are composed of tiny particles that are arranged far apart from each other
Gases are composed of individual atoms such as in the element neon or molecules such as in the element oxygen
Objectives #1-4: Introduction to the Kinetic Theory of Gases
Collisions that may occur between gas particles are elastic with no net loss of energy
Gas particles are in constant, random motion
No attractive forces exist between gas particles
The temperature of a gas depends on the average kinetic energy of the particles
Objectives #1-4: Introduction to the Kinetic Theory of Gases
II. The Kinetic Theory and its Implications on the Properties of GasesThe temperature of a gas is directly related to its kinetic energyThe temperature at which no kinetic energy is present is called absolute zero; this temperature is 0 on the Kelvin scale and -273o on the Celsius scaleGases are able to expand freely due to their random motion and the lack of attractive forces between the particles of a gas
Objectives #1-4: Introduction to the Kinetic Theory of Gases
The density of gases is generally less than other substances due to the ability of gases to move freely
Gases can be compressed due to the great distances between gas particles
The ability of gases to spread out spontaneously or diffuse is due to the rapid motion of gas particles; particles with less mass and faster velocities spread out at a faster rate than heavier, slower gas particles (This is why you smell cookies baking! )
Objectives #1-4: Introduction to the Kinetic Theory of Gases
Due to their small size, gas particles can be made to pass through small openings; this characteristic of effusion depends on the velocity and molar mass of the gas particles present; the rate of effusion is directly proportional to the velocity of the gas particles and inversely proportional to the molar mass of the gas particles
When gas particles collide against a surface they exhibit pressure; which is defined as the force per unit area
Objectives #5-8: Relationships Among Gas Characteristics
Common Units of Pressure:
Unit Symbol Standard Value
Pascal/Kilopascal Pa/kPa 101325/101.3
Millimeters of Hg mm Hg 760
Atmospheres atm 1
Torr torr 760
Objectives #5-8: Relationships among Gas Characteristics
Pressure Conversion Examples:• Convert 2.5 atm to mmHg
2.5 atm x 760 mmHg = 1900 mmHg
1 atm• Convert 300. Pa to atm
300. Pa x 1 atm_____ = .00296 atm 101325 Pa
Objectives #5-8: Relationships among Gas Characteristics
Dalton’s Law of Partial PressureThe total pressure of a mixture of gases is equal
to the sum of the partial pressures of each of the individual gases in the mixture
PT = P1 + P2 + P3 + ….This concept must be kept in mind when gases are
collected over water in the laboratory; the vapor pressure of water must be subtracted from the measured pressure of the gas in order to obtain the true pressure of the gas being collected
Pgas = Patm - Pwater
Objectives #5-8: Relationships among Gas Characteristics
The vapor pressure due to water increases with increasing temperature (see attached chart)
Molar Volumeat standard temperature and pressure
(STP), which are defined as O o C and 1 atm, 1 mole of any gas occupies 22.4 L; this is referred to as molar volume
Objectives #5-8: Relationships among Gas Characteristics
Relationships Among Gas CharacteristicsA. Amount of Gas vs. Pressure
(assumes temperature and volume are held constant)
What is happening at the particle level:
How can you tell the can is full?
Gases expand to fit the space, so increasing the particles makes the can heavier !
Objectives #5-8: Relationships among Gas Characteristics
As the amount of gas in a container increases, the pressure increases
This illustrates a direct relationship between the amt. of the gas and the pressure of the gas
Empty can = less pressure, so the can is lighter weight
Objectives #5-8: Relationships Among Gas Characteristics
B. Pressure vs. Volume of Gases (assumes constant temperature)
What is happening at the particle level:
Why are balloons “over-filled” with helium?
Objectives #5-8: Relationships among Gas Characteristics
As the pressure of a fixed amount of gas increases, its volume decreases
This illustrates an inverse relationship
Objectives #5-8: Relationships among Gas Characteristics
C. Temperature vs. Volume (assumes constant pressure)
What is happening at the particle level: Why does tires tend to
look like they are low in air on a cold winter morning?
Cold night temperatures result in slower movement of the gas particles, so the volume of the gas seems to decrease. As the car moves, friction of the tire on the road increases the temp of the air in the tire.
Objectives #5-8: Relationships Among Gas Characteristics
As the Kelvin temperature of a fixed amount of gas increases, its volume increases
This illustrates a direct relationship:
Objectives #5-8: Relationships Among Gas Characteristics
D.Temperature vs. Pressure (assuming constant volume)
What is happening at the particle level: Keeping & transporting
unopened pop cans in the car in the summer is NOT advisable. Why ?
Increasing the temperature in the car increases the pressure in the can. If the pressure of the gas is greater than the pressure exerted by the wall of the can, the can will explode!
Objectives #5-8: Relationships Among Gas Characteristics
As the Kelvin temperature of a fixed amount of gas increases, its pressure increases
This illustrates a direct relationship
Objectives #5-8: Relationships Among Gas Characteristics
Ideal vs. Real GasesIdeal gases always follow the kinetic
theory under any conditionsReal gas particles do have
attractive forces among each otherReal gases no longer act as ideal
gases under conditions of high pressure and extremely low temperature
Objective #9 Solving problems Involving the Gas Laws
*the Gas Laws*the Gas Laws are mathematical
formulas based on the relationships discussed in the previous section of notes
P1V1 / T1 = P2V2 / T2 (Combined Gas Law)
PV = nRT (Ideal Gas Law)I. Combined Gas Law*examples:
1. To what pressure must a gas be compressed to get it into a 9.00 L tank if it occupies 90.0 L at 1.00 atm?
Can a variable be eliminated?P1V1 = P2V2
P1V1 / V2 = P2
(1.00 atm) (90.0 L) / 9.00 L = P2
10.0 atm = P2
2. A container with a movable piston contains .89 L of methane gas at 100.50C. If the temperature of the gas drops to 11.3oC, what is the new volume of the gas?
K = C + 273 = 100.5oC + 273 = 374 K
K = C + 273 = 11.3oC + 273 = 284 KCan a variable be eliminated?V1 / T1 = V2 / T2
V1T2 = T1V2
V1T2 / T1 = V2
(.89 L) (284 K) / 374 K = V2
.676 L = V2
3. A sample of gas occupies a volume of 5.0 L at a pressure of 650. torr and a temperature of 24oC . We want to put the gas in a 100. ml container that can only withstand a pressure of 3.0 atm. What temperature must be maintained so that the container does not explode?
5.0 L = 5000 ml650. torr = .855 atm24oC = 297 KP1V1 / T1 = P2V2 / T2
Can a variable be eliminated?P1V1T2 = T1P2V2
T2 = T1P2V2 / P1V1
T2 = T1P2V2 / P1V1
T2 = (297 K) (3.0 atm) (100. ml) /
= (.855 atm) (5000 ml) = 21 K4. A sample of gas occupies 2.00 L at
STP. What volume will it occupy at 27oC and 200. mmHg?
P1V1 / T1 = P2V2 / T2
Can a variable be eliminated?P1V1T2 = T1P2V2
P1V1T2 / T1P2 = V2
(760 mm Hg) (2.00 L) (300 K) /(273 K) (200. mm Hg) = V2
8.35 L = V2
II. Additional Problems*Recall that at STP conditions, 1 mole
of any gas = 22.4 L*examples:1. Calculate the volume of .55000
moles of gas at STP..55000 moles X 22.4 L / 1 mole =
12.320 L
2. Calculate the moles of gas contained in 350 L of gas at STP.
350 L X 1 mole / 22.4 L = 16 moles3. Calculate the mass in grams of 3.50
L of chlorine gas.3.50 L X 71.0 g / 22.4 L = 11.1 g
Objective #10 The Ideal Gas Law and Gas Stoichiometry
PV=nRT*ideal gas constant:R = PV / ntR = ( 1 atm) (22.4 L) / (1 mole) (273K) = .0821 L . atm. / mole.K*examples:1. What is the temperature of a .65 L sample
of fluorine gas at 620. torr which contains 1.3 mol fluorine?
T = PV / nR
*examples:1. What is the temperature of a .65 L
sample of fluorine gas at 620. torr which contains 1.3 mol fluorine?
T = PV / nR = (620 torr / 760 torr) (.65 L) / (1.3 mol) (.0821 L.atm / mol.K) = 5.0 K
2. A 25.0 gram sample of argon gas is placed inside a container with a volume of 10.0 L at a temperature of 65oC. What is the pressure of argon at this temperature?
PV=nRTP = nRT / V = (25.0 g / 39.9 g) (.0821) (338 K) / 10.0 L = 1.74 atm.
*there are two types of gas stochiometry problems:
at STPnon STP*examples:
1. Calculate the volume of hydrogen gas that can be produced from the reaction of 5.00 g of zinc reacted in an excess of hydrochloric acid. Assume STP conditions.
Zn + 2HCl --› H2 + ZnCl25.00 g Zn X 1 mole Zn / 65.4 g Zn X1 mole H2 / 1 mole Zn X 22.4 L / 1 mole H2
= 1.71 L
2. Calculate the volume in liters of oxygen gas that can be produced from the decomposition of 3.50 X 1024 formula units of potassium chlorate. Assume STP conditions.
2KClO3 --› 2KCl + 3O2
3.50 X 1024 formula units KClO3 X
1 mole KClO3 / 6.02 X 1023 f. units X
3 mole O2 / 2 mole KClO3 X 22.4 L / 1 mole =
195 L
3. Calculate the volume of hydrogen produced at 1.50 atm and 19oC by the reaction of 26.5 g of calcium metal with excess water. (Ignore the vapor pressure of water)
Ca + 2H2O --› H2 + Ca(OH)2
*use amount of given reactant and stoichiometry to determine moles of gas desired in problem:
26.5 g Ca X 1 mole Ca / 40.1 g Ca X1 mole H2 / 1 mole Ca = .661 mole H2
*use moles of gas found in ideal gas law to calculate volume of gas:
PV=nRTV = nRT / P = (.661 moles) (.0821) (292 K) / 1.50
L = 10.6 L
4. Calculate the volume of chlorine gas produced at 1.25 atm at 25oC from the reaction of 5.00 g of sodium chloride and an excess of fluorine.
2NaCl + F2 --› 2NaF + Cl25.00 g NaCl X 1 mole NaCl / 58.5 g
NaCl X1 mole Cl2 / 2 mole NaCl = .0427 mole
Cl2
PV = nRTV = nRT / P = (.0427 mole) (.0821) (298 K) /1.25
atm = .836 L